# Help with this Diophantine equation

Note: This question was posted in error, and should be closed as no longer relevant. The correct question is posted at Help with this system of Diophantine equations (End of note)

For a research problem that I'm working on, I need to solve this Diophantine equation:-

$a^3+b^3+c^3-3d=-83449$, where $a,b,c,d>0$ are all DISTINCT positive integers and$a,b,c∉${$2,9,15,16,33,34$}.

How does one go about solving this? Is brute-force the only possible way? Or could there be a case that no integer solutions exist for this equation?

Also, are there any online computing engines, that allow me to set constraints, and solve Diophantine equations of this sort?

Any and all help is appreciated! Thanks!

-
Your conditions imply a is less than 70, and from this you can get small bounds on b and c. This is well within a brute force search with a laptop, and you can use digital considerations and perhaps a table of cubes to do much of the search by hand. Gerhard "Ask Me About System Design" Paseman, 2013.05.21 – Gerhard Paseman May 21 '13 at 13:49
@Gerhard: Why do the conditions imply that $a < 70$? – Stefan Kohl May 21 '13 at 13:55
Stefan, I made the mistake of assuming b and c were both larger than a. I see now that there is no bound on $a^3$ when it is larger than one of the cubes. There are some restrictions mod 3 on b-a and c-a, but they don't seem as useful now. Gerhard "Missed It By A Lot" Paseman, 2013.05.21 – Gerhard Paseman May 21 '13 at 14:50
Thank you guys, for the replies! But sorry to say, my original equation was at fault. I've edited the question now. – Anton Schigur May 21 '13 at 15:15
This question (in its original form) was posted at math.stackexchange.com/questions/398112/… -- @Jobin, please do not cross-post the same question at both sites -- it can result in unnecessary duplication of effort. (It is sometimes OK to do so, if time has passed and you haven't gotten an answer at math.stackexchange, but even then you should say so and include the link.) – Barry Cipra May 21 '13 at 15:59

For $0 < a \le3966887$ solutions are $(9419, 10418, 8146),(69167, 10776, 87090)$ and (added) $(3966887, 2434179, 4797573)$.

Here is an idea for searching. Loop $a$ from $1$ to certain bound.

You have to solve $x^3 + y^3 = C + 2 a^3 = N$. This is easy to solve if $N$ can be factored since $x^3+y^3$ factors nicely.

You have to solve $a^3+b^3+c^3 + 83449 = 3 d$

Just pick "random" $a,b,c$ such the the lhs is divisible by $3$ like $(300,301,304)$ and $d=27482938$

Here is a pari/gp script which found the solutions.

{
jobin1()=
th=thueinit(x^3+1,1);
C=36650;
for(a=1,10^5,
A=C+2*a^3;
v=thue(th,A);
if(v == [],next);
print([a,v]);
);
}
jobin1()
-
I just realized I'd made a huge mistake in my assumptions for the problem. Subsequent calculations reveal my required Diophantine equation to be a^3 + b^3 + c^3 - 3d = -83449. Here, 'd' is also a positive integer, but NOT subjected to the constraint that d be in {2,9,15,16,33,34}. Could you help solve this, please? – Anton Schigur May 21 '13 at 15:07
Do you mean "that $d$ not be in"? – Joel Reyes Noche May 21 '13 at 15:30
I'm terribly sorry for troubling you again, joro. I'd committed a blatant "freshman" error again. I've newly posted the question I'd actually meant to ask here:- mathoverflow.net/questions/131353/… – Anton Schigur May 21 '13 at 16:03

Sorry for a long comment. Joro has given a nice answer already. Writing $36650=b^3+c^3-a^3-a^3$, the question is related to the problem which numbers can be represented by the sum of $4$ signed cubes. A result of Demjanenko says that all numbers not of the form $9n \pm 4$ are representable as a sum of four signed cubes. Indeed, all integers $n\le 10^7$ have such a representation, and for $n$ sufficently large the representation also exists (see the artcle Kenji Koyama, On searching for solutions of the Diophantine equation $x^3 + y^3 + 2z^3 = n$ , Math. Comput. 69 (2000).

EDIT: the new equation seems to be $a^3+b^3+c^3=n=3d-83449$. The conjecture is that this has solutions if and only if $n$ is not of the form $9k\pm 4$.

-
Not sure if this is correct, check the counterexamples in my answer. – joro May 21 '13 at 15:03
Sorry guys for the trouble, but my question was in error. I've edited it accordingly now. – Anton Schigur May 21 '13 at 15:18
From matematicas.unal.edu.co/boletin/Archivos/2009-II/16.2_3.pdf ("On partitions into four cubes" by Moreno and Palma): Demjanenko [7] proved that every number $n \nequiv \pm4 \mod 9$ can be expressed as the sum of four positive or negative cubes: $x^3 + y^3 + z^3 + w^3$. – Barry Cipra May 21 '13 at 15:36
sorry, that "\nequiv" was meant to be a $\not\equiv$. – Barry Cipra May 21 '13 at 15:38

Let a,b,c satisfy the restrictions given, as well as $1 + a + b + c$ is a multiple of $3$. Then $83449 + a^3 + b^3 + c^3$ is also a multiple of 3, and then $d$ can be chosen to be $1/3$ of the last quantity.

Gerhard "3D Makes It So Easy" Paseman, 2013.05.21

-